Updated: Following Michael's suggestion, I rephrase the question slightly.

Given a locally convex (Hausdorff) topological vector space (LCTVS), when is it isomorphic to a subspace of some function space $Y^X$ (equipped with the product topology), where $Y$ is, say, some Banach space (if it helps simplify things, can assume $Y=\mathbb{C}$, the complex field), and $X$ is some set. We are free to choose X and Y.

If not all LCTVS have this property, then what kind of conditions do we need?

Now, perhaps you should tell us what you found when you looked in a textbook on locally convex spaces...
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Gerald EdgarApr 30 '12 at 3:13

I think I got the answer: It seems to be always true as long as the LCTVS is complete (which really is necessary for $Y^X$ is complete when $Y$ is so). It is in the book "Functional Analysis: Theory and Applications" by Robert E. Edwards, Ex. 6.17.
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yaoliangApr 30 '12 at 4:57

2 Answers
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If I interpret the question correctly, Yaoliang would like to know which LCTVS are isomorphic to $\mathbb{C}^X$, where $X$ is a set (no topology), and $\mathbb{C}^X$ is given the product topology. If this is so, the answer is: very few. Actually, spaces of this form are fully determined by the cardinality of $X$.

Just to make an example: no infinite dimensional normed space can be isomorphic to a space of this form. Indeed, in this case the set $X$ would have to be infinite, but then every neighborhood of $0$ in $\mathbb{C}^X$ would contain a proper vector subspace, and this is never true for normed spaces.

Thanks Alberto! It is a nice counterexample. Can you tell me more why such spaces are fully determined by the cardinality of X? Let us assume the cardinality is infinite (otherwise not interesting). Note that we are free to choose X when realizing the LCTVS as a function space. I also don't mind replacing the complex field as some Banach space (or even more general, if it helps), i.e., the function space is some Banach space valued function over some set X.
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yaoliangApr 29 '12 at 17:18

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A bijection between $X$ and $Y$ induces an isomorphism (of TVS) between $\mathbb{C}^X$ and $\mathbb{C}^Y$, so only the cardinality of $X$ matters. Notice also that in your class of spaces $\mathbb{C}^X$ there is only one infinite dimensional separable space, the one which you obtain for $X=\mathbb{N}$. The problem is that you are considering ALL functions from $X$ to $\mathbb{C}$ (and if you replace $\mathbb{C}$ by a larger vector spaces things get even worse). If you want to represent interesting TVS you should put more structure on $X$ and restrict the class of functions.
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Alberto AbbondandoloApr 29 '12 at 17:54

A potentially more interesting question might be which LCTVS are isomorphic to a subspace of $C^X$ for some X.
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Michael RenardyApr 29 '12 at 20:25

Yes, I thought subspaces of $C^X$ is also $C^{X'}$ for a different $X'$ :). I should ask differently.
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yaoliangApr 30 '12 at 2:21

Yes, every LCTVS can be realized as a function space: every LCTVS is isomorphic to the dual of its dual space equipped with the weak* topology. This is a consequence of a version of the Hahn-Banach theorem which states that in a real LCTVS two disjoint closed convex sets, one of which is compact, can be strictly separated. See Section IV.3 of A Course in Functional Analysis by Conway.

... but why is the original LC topology equal to the product topology of the function space ???
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Gerald EdgarApr 29 '12 at 12:35

You're right, I should have read the question more carefully. Not every LC topology is induced by a family of linear functionals --- for instance, if a topology is induced by a family of linear functionals then every open neighborhood of zero contains a finite codimension subspace. I guess the answer is that you can realize a LCTVS in this sense as a function space if and only if its original topology equals its weak topology. I'm not sure there's going to be any more concrete answer than that.
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Nik WeaverApr 29 '12 at 13:34

You can always realise it as a subspace of the space of continuous functions on the dual by using the topology of uniform convergence on the equicontinuous sets of the latter, raher than the weak topology.
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blackburneJul 24 '14 at 7:17